General Ehrenfest Theorem applied to N-qubit system operator

Question

Please advise if the following short calculation of the derivative of the expectation value of an all spin Pauli-$\hat{Y}$ operator (acting on a $N$-qubit system) is consistent:

The general Ehrenfest theorem states $$\frac{d }{d t}\langle \hat{S}(t) \rangle = \bigg\langle \frac{\partial \hat{S}(t)}{\partial t} \bigg\rangle + \frac{1}{i \hbar}\langle [\hat{S}(t),\hat{H}]\rangle~~~~~~~~(*),$$ for some operator $\hat{S}$. Consider the expectation value of the all Pauli-spin $Y$ measurement operator $\hat{S}= \underbrace{\hat{Y}\otimes \cdot \cdot \cdot \otimes \hat{Y}}_{N}$ (in the Heisenberg picture) $$\big\langle\hat{S}(t)\big\rangle = \big\langle e^{i \frac{t}{2} \hat{X_{1}}}\hat{Y_{1}}e^{-i \frac{t}{2} \hat{X_{1}}}\otimes\cdot\cdot\cdot\otimes e^{i \frac{t}{2} \hat{X_{N}}}\hat{Y_{N}}e^{-i \frac{t}{2} \hat{X_{N}}} \big\rangle$$ acting on an $N$-qubit system, where the Hamiltonian is given by $\hat{H} := \sum_{l=1}^{N}\hat{X}_{l}$. Would I be correct in stating that this implies that the expectation value terms $(*)$ are given by \begin{align}\frac{\partial}{\partial t}[e^{i \frac{t}{2} \hat{X_{1}}}\hat{Y_{1}}e^{-i \frac{t}{2} \hat{X_{1}}}\otimes\cdot\cdot\cdot\otimes e^{i \frac{t}{2} \hat{X_{N}}}\hat{Y_{N}}e^{-i \frac{t}{2} \hat{X_{N}}}] = \frac{\partial}{\partial t}[e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}]&\\=\underbrace{[i\hat{X}e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}]+\cdot\cdot\cdot +[e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes i\hat{X}e^{i t \hat{X_{N}}}\hat{Y}_{N}] }_{N}\end{align}

using the anti-commutation of Pauli $\hat{X}$ and $\hat{Y}$ operators and then the product rule of differentiation of a tensor product. For the second term we get \begin{align}[\hat{S}(t), \hat{H}]= [e^{i t \hat{X_{1}}}\hat{Y_{1}}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y_{N}},\sum_{l}\hat{X}_{l}] = \underbrace{[e^{i t \hat{X_{1}}}\hat{Y}_{1}\hat{X_1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y_{N}}]+\cdot\cdot\cdot +[e^{i t \hat{X_{1}}}\hat{Y_{1}}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}\hat{X_N}] }_{N}. \end{align}

Thanks for any assistance.

Answer 1

The expression you have for $\langle S(t)\rangle$ on the top is already the time dependence - in this case there's no need to calculate anything beyond that. If, however, you want to see how the calculation turns out when doing it explicitly, it is:

First of all, because there are no interactions, it's much easier to look at a single qubit at a time and in the end take the tensor product. So let's do that: $$ \frac{d}{dt}\langle Y\rangle = \bigg\langle \frac{\partial Y(t)}{\partial t} \bigg\rangle -i\langle [Y,H]\rangle $$ Now, $Y$ has no explicit time dependence, so its partial derivative w.r.t time is zero. Secondly, $$ -i[Y(t),H]=-i[Y(t),X]=-2Z(t) $$ so (no need for the averaging anymore, we're basically working in the Heisenberg picture): $$ \frac{dY(t)}{dt}=-2Z(t) $$ Similarly, $$ \frac{dZ(t)}{dt} =2Y(t). $$ Since $Y(0)=Y$ and $Z(0)=Z$, the solution is: $$ Y(t)=Y\cos 2t-Z\sin2 t $$ And the solution for the full many qubit system is the tensor product of this. From it, you just need to calculate the expectation value.

Please let me know if this answers your question!